Results 1 to 5 of 5

Math Help - Epsilon-proof as x->-infinity currently defeating me

  1. #1
    Junior Member
    Joined
    Mar 2009
    Posts
    58

    Epsilon-proof as x->-infinity currently defeating me

    I think Epsilon-delta proofs and I should spend some quality time together.

    The problem is I need to prove that:

    \lim_{x \to -\infty} a^x = 0

    So, I need to show that, for every \epsilon > 0, there is a \delta such that |f(x) - L| < \epsilon whenever x < \delta.

    So, given \epsilon > 0, I need to find \delta such that:

    |a^x - 0| < \epsilon whenever x < \delta

    And here's where I am stuck. I think I'll have to take a log in there somewhere. Wish my textbook wasn't so scarce on examples (Stewart's Calculus 4th edition). I sort of get limits where x->c, but I'm totally lost on the ones with infinities, and the differences between handling + and - infinities. Can't find any examples online of ones as x-> -infinity, which surely might help. Nor are the two videos at Khan Academy useful in this case.

    I bet you guys get a lot of these. Surprisingly confusing for what seems a relatively simple concept, at first. Help, hints, pointers, nudges and taunts all appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    If you think of \delta as a large negative number, and getting more and more negative, you'll be on the right track; also, you have to assume that a>0, right? Otherwise, you have complex numbers floating around.

    How can you simplify |a^{x}-0|<\epsilon?

    If you can find a \delta=\delta(\epsilon) that works, you'll be done. How could you do that?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2009
    Posts
    58
    Quote Originally Posted by Ackbeet View Post
    If you think of \delta as a large negative number, and getting more and more negative, you'll be on the right track; also, you have to assume that a>0, right? Otherwise, you have complex numbers floating around.

    How can you simplify |a^{x}-0|<\epsilon?
    Well, first and obviously, |a^{x} - 0| < \epsilon \Rightarrow |a^{x}| < \epsilon.

    Then, I can assume that a^{x} > 0 so I have just:

    a^{x} < \epsilon whenever x < \epsilon.

    Quote Originally Posted by Ackbeet View Post
    If you can find a \delta=\delta(\epsilon) that works, you'll be done. How could you do that?
    We want:

    a^{x} < \epsilon whenever x < \epsilon

    Or:

    log_{a}(a^{x}) = log_{a}(\epsilon) \Rightarrow x = log_{a}(\epsilon)

    So I should try \delta = log_{a}(\epsilon).

    Given \epsilon > 0, we choose \delta = log_{a}(\epsilon). Let x < \epsilon. Then

    |a^{x} - 0| = a^{x} < a^{\delta} = a^{log_{a}(\epsilon)} = \epsilon

    Thus

    |a^{x} - 0| < \epsilon

    Is that right? Egads, I think I just did it. Hey, I think I smell burnt toast!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    Quote Originally Posted by Grep View Post
    I think Epsilon-delta proofs and I should spend some quality time together.

    The problem is I need to prove that:

    \lim_{x \to -\infty} a^x = 0

    ...
    ... that's true if  a >1... of course...



    Merry Christmas from Italy

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2009
    Posts
    58
    Quote Originally Posted by chisigma View Post
    ... that's true if  a >1... of course...
    Good point, the problem description says "Let a > 1", which I should have stated. Apologies for leaving out an important part of the problem description.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 15
    Last Post: June 8th 2011, 11:13 AM
  2. Replies: 3
    Last Post: February 1st 2011, 12:36 PM
  3. Solving Delta Epsilon Proof (Given Epsilon)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 15th 2010, 03:42 PM
  4. help with epsilon delta proof
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 19th 2010, 05:12 PM
  5. Epsilon Delta Proof with Infinity
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 20th 2009, 07:19 AM

Search Tags


/mathhelpforum @mathhelpforum